Expression for shock wave is given as : Ratio of change in flow to change in density.
Given are the q ( flow ) and u ( speed ) values, and we know that k (density ) = q / v. (By Greenshield's model, linear relarionship)
So,
Speed of shock wave AB = ( 1543 - 1750 ) / ( ( 1543/18 ) - ( 1750/35 ) ) = 5.79 mi/h (backward direction)
Speed of shock wave BC = ( 1714 - 1543 ) / ( ( 1714/30 ) - ( 1543/18 ) ) = 5.98 mi/h (backward direction)
Now, the queue will be behind the school zone.
Hence, queue length, N = (1750/35) * ( 35 - ( 5.79 ) ) * ( 20/60 ) = 1750 * 29.21 / ( 3 * 35 ) = 486.83 or 487 vehicles
This queue will dissipate, when zone B stops operating, hence,
487 = (1543/18) * ( 18 - 5.98 ) * t
On solving for t, we get, t = 0.472 hours or t = 28.3 min.
Hence, it takes 28.3 minutes to dissipate the queue.
Studies have shown that the traffic flow on a two-lane road adjacent to a school eenshields...
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